|
|
Re: No Identity Bijection for Omega
Posted:
Sep 4, 2007 12:21 PM
|
|
On Sep 3, 10:28 pm, logic...@comcast.net wrote:
> Let B be the identity bijection of w+1 = w U {w} .
> B is an ordered set of ordered pairs.
>
> B = < <0,0>, <1,1>, <2,2>, <3,3>, ..., <w,w> >
>
And clearly the identity bijection for w is
A = < <0,0>, <1,1>, <2,2>, <3,3>, ... >
So your proof that A does not exist must be flawed.
Hint. A does not have a last element.
> For each element, B_i, let C_i be the set of
> all B_i that appear in the bijection before B_i.
> Let C_i also includes B_i. (Skeptics - this is the one to attack.)
>
> Let D_i be the set that C_i is a bijection for.
> (D_i is the set of all natural numbers that appear in C_i).
>
> B_0 = <0,0>
> B_1 = <1,1>
> B_2 = <2,2>
> ...
>
> B_w = <w,w>
>
> C_0 = < <0,0> >
> C_1 = < <0,0>, <1,1> >
> C_2 = < <0,0>, <1,1>, <2,2> >
> ...
> C_w = < <0,0>, <1,1>, <2,2>, ..., <w,w> >
>
> D_0 = {0}
> D_1 = {0,1}
> D_2 = {0,1,2}
> ...
> D_w = {0,1,2,...,w}
>
> Assume D_k = w for some k.
> D_k = {0,1,2,...,k} and k must be the largest natural number.
> Therefore, no D_k = w.
>
> Assume there exists a C_k that is a bijection for w.
> Then there exists D_k.
>
> Assume there exists an identity bijection, A, for w.
> A must be an proper subset of B.
> A must also be an initial segment of B.
Ok to this point.
> If A is an initial segment of B then A = C_k for some k.
No. This is simply false.
If A is an initial segment of B *that has a last element*
then A = C_k for some k.
If A is an initial segment of B *that does not have a last element*
then A does not equal C_k for any k.
- William Hughes
|
|
